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bezimeni [28]
3 years ago
8

Find the area inside the semicircle and out side the triangle in terms of ń

Mathematics
1 answer:
Elan Coil [88]3 years ago
6 0

Answer:

π 169 / 2 - 120

Step-by-step explanation:

I'm assuming you are asking question about 24.

So first find radius of semicircle:

a^2+b^2=c^2

10^2+24^2=c^2

c=26

radius 26/2 = 13

Then let's calculate area of semicircle

area of circle = π*r^2

area of semicircle = π*r^2 / 2

area of given semi circle = π 169 / 2

Then let's calculate area of triangle

area of triangle = 10 * 24 / 2 = 120

Area of the shaded region =  π 169 / 2 - 120

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HELLPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
Setler79 [48]

Answer:

65 square units

Step-by-step explanation:

The area of a triangle is: height times base divided by two.

So the area of one of the triangles is 10 square units.

multiply this times 4 to get the total area of the triangles

then add the area of the square which is 5 times 5

25 plus 40 = 65 square units

PLZ MARK BRAINLIEST

3 0
3 years ago
Read 2 more answers
Please help with this one
tino4ka555 [31]

The answer will be A-1188 in²

3 0
2 years ago
Read 2 more answers
There were 29 students available for the woodwind section of the school orchestra. 11 students could play the flute, 15 could pl
Dafna11 [192]

Answer:

a. The number of students who can play all three instruments = 2 students

b. The number of students who can play only the saxophone is 0

c. The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only one of the clarinet, saxophone, or flute = 4

Step-by-step explanation:

The total number of students available = 29

The number of students that can play flute = 11 students

The number of students that can play clarinet = 15 students

The number of students that can play saxophone = 12 students

The number of students that can play flute and saxophone = 4 students

The number of students that can play flute and clarinet = 4 students

The number of students that can play clarinet and saxophone = 6 students

Let the number of students who could play flute = n(F) = 11

The number of students who could play clarinet = n(C) = 15

The number of students who could play saxophone = n(S) = 12

We have;

a. Total = n(F) + n(C) + n(S) - n(F∩C) - n(F∩S) - n(C∩S) + n(F∩C∩S) + n(non)

Therefore, we have;

29 = 11 + 15 + 12 - 4 - 4 - 6 + n(F∩C∩S) + 3

29 = 24 + n(F∩C∩S) + 3

n(F∩C∩S) = 29 - (24 + 3) = 2

The number of students who can play all = 2

b. The number of students who can play only the saxophone = n(S) - n(F∩S) - n(C∩S) - n(F∩C∩S)

The number of students who can play only the saxophone = 12 - 4 - 6 - 2 = 0

The number of students who can play only the saxophone = 0

c. The number of students who can play the saxophone and the clarinet but not the flute = n(C∩S) - n(F∩C∩S) = 6 - 2 = 4

The number of students who can play the saxophone and the clarinet but not the flute = 4 students

d. The number of students who can play only the saxophone = 0

The number of students who can play only the clarinet = n(C) - n(F∩C) - n(C∩S) - n(F∩C∩S) = 15 - 4 - 6 - 2 = 3

The number of students who can play only the clarinet = 3

The number of students who can play only the flute = n(F) - n(F∩C) - n(F∩S) - n(F∩C∩S) = 11 - 4 - 4 - 2 = 1

The number of students who can play only the flute = 1

Therefore, the number of students who can play only one of the clarinet, saxophone, or flute = 1 + 3 + 0 = 4

The number of students who can play only one of the clarinet, saxophone, or flute = 4.

6 0
3 years ago
Write a multiplication fact that goes with this division fact.<br><br> 30 ÷ 5 = 6
34kurt

5 times 6 = 30 hope this helped

8 0
3 years ago
Read 2 more answers
Paul has 12 coins in his pocket, consisting entirely of dimes and quarters. If he has a total of 195 cents in coins, how many co
anyanavicka [17]

Let x be the number of dimes

and y be the number of quarters

Total number of coins = 12

So number of dimes + number of quarters = total number of coins

x + y = 12 ---------> equation 1

We know, 1 Quarter = 25 cents, 1 Dime = 10 cents, total cents = 196

y quarters = 25y cents , x dimes = 10x cents

x dimes + y quarters = 195 cents

10x + 25y = 195 -----------> equation 2

Now solve both equations

x + y =12, we solve for y by subtracting x on both sides

y = 12-x

Now substitute y in second equation

10x + 25(12-x) = 195

10x + 300 -25x = 195 (combine like terms)

-15x + 300 = 195 (subtract 300 on both sides)

-15x = -105 ( divide by -15)

x=7

y= 12-x

y=12-7 = 5

The number of dimes = 7

Number of quarters = 5






8 0
3 years ago
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